Question 1

A disk has a moment of inertia of 3.0 *10^-⁴ kg·m² and rotates with an angular speed of 3.5 rad/sec. What net torque must be applied to bring it to rest within 3.7 s? A) 4.6 * 10^-⁴ N·m B) 7.5 * 10^-⁴ N·m C) 3.5 *10^-⁴ N·m D) 2.8 * 10^-⁴ N·m

Accepted Answer

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Question 2

A bucket in an old well is hoisted upward by a rope which winds up on a cylinder having a radius of 0.050 m. How many rev/s must the cylinder turn if the bucket is raised at a speed of 0.20 m/s?
A) 3.0 rev/s
B) 1.5 rev/s
C) 0.48 rev/s
D) 0.64 rev/s

Accepted Answer

The answer of A bucket in an old well is...

Question 3

A 0.40-m radius automobile tire accelerates from rest at a constant 2.0 rad/s². What is the centripetal acceleration of a point on the outer edge of the tire after 5.0 s? A) 4.0 m/s² B) 40 m/s² C) 44 m/s² D) 400 m/s²

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Question 4

A ventilation fan with a moment of inertia of 0.034 kg·m² has a net torque of 0.14 N·m applied to it. What angular acceleration does it experience? A) 5.3 rad/s² B) 4.1 rad/s² C) 3.2 rad/s² D) 0.31 rad/s²

Accepted Answer

The equation used for this problem is:
τ = Iα
where τ is the net torque, I is the moment of inertia, and α is the angular acceleration.

Rearranging the equation, we get:
α = τ/I

Plugging in the given values, we get:
α = 0.14 N·m / 0.034 kg·m^2
α = 4.1 rad/s^2

Therefore, the angular acceleration experienced by the ventilation fan is 4.1 rad/s^2. The best choice for this answer is B.

Question 5

A ventilation fan has blades 0.25 m in radius rotating at 30 rpm. What is the tangential speed of each blade tip?
A) 0.02 m/s
B) 0.52 m/s
C) 0.79 m/s
D) 20 m/s

Accepted Answer

First, convert the rpm to radians per second:
30 rpm = 30/60 = 0.5 rps
Now, the tangential speed (v) can be found using the formula:
v = rω
where r is the radius and ω is the angular velocity in radians per second.
So, v = 0.25 x 0.5 x 2π = 0.79 m/s (rounded to two decimal places)
Therefore, the answer is C, 0.79 m/s.

Question 6

If a nonzero torque is applied to an object, that object will experience:
A) a constant angular speed.
B) an angular acceleration.
C) a constant moment of inertia.
D) an increasing moment of inertia.

Accepted Answer

According to Newton's second law for rotation, torque (τ) is equal to the moment of inertia (I) multiplied by the angular acceleration (α): τ = Iα. Therefore, if a nonzero torque is applied to an object, it will experience an angular acceleration (α) and its angular speed will change over time. The moment of inertia (I) of the object will only change if its shape or mass distribution changes.

Question 7

A solid sphere of radius R is placed on an incline that makes an angle  with the horizontal. The sphere rolls down the incline without slipping. What is its acceleration?
A) 
B) 
C) 
D)

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Question 8

An automobile accelerates from zero to 30 m/s in 6.0 s. The wheels have a diameter of 1.0 m. What is the average angular acceleration of each wheel? A) 5.0 rad/s² B) 10 rad/s² C) 25 rad/s² D) 40 rad/s²

Accepted Answer

First, we need to find the angular velocity of each wheel at the end of the 6.0 s acceleration. We can use the formula:

v = ωr

where v is the linear velocity (30 m/s), r is the radius of the wheel (0.5 m) and ω is the angular velocity.

ω = v/r = 30 m/s ÷ 0.5 m = 60 rad/s

Then, we can use the formula for average angular acceleration:

α = Δω/Δt

where Δω is the change in angular velocity and Δt is the time it took to change.

Δω = 60 rad/s - 0 rad/s = 60 rad/s

Δt = 6.0 s

α = 60 rad/s ÷ 6.0 s = 10 rad/s^2

Therefore, the average angular acceleration of each wheel is 10 rad/s^2, which corresponds to choice B.

Question 9

A solid cylinder (I = MR²/2) has a string wrapped around it many times. When I release the cylinder, holding on to the string, the cylinder falls and spins as the string unwinds. What is the downward acceleration of the cylinder as it falls? A) 0 B) 4.9 m/s² C) 6.5 m/s² D) 9.8 m/s²

Accepted Answer

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Question 10

A bucket of water with total mass 30 kg is attached to a rope, which in turn is wound around a 0.050 m radius cylinder at the top of a well. A crank with a turning radius of 0.25 m is attached to the end of the cylinder and the moment of inertia of cylinder and crank is 0.12 kg*m². If the bucket is raised to the top of the well and released, what is the acceleration of the bucket as it falls toward the bottom of the well? (Assume rope's mass is negligible, that cylinder turns on frictionless bearings and that g = 9.8 m/s².) A) 3.2 m/s² B) 3.8 m/s² C) 7.4 m/s² D) 9.8 m/s²

Accepted Answer

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Question 11

A baseball pitcher loosens up his pitching arm. He tosses a 0.15-kg ball using only the rotation of his forearm, 0.32 m in length, to accelerate the ball. If the ball starts at rest and is released with a speed of 24 m/s in a time of 0.40 s, what torque is applied to the ball while being held by the pitcher's hand to produce the angular acceleration?
A) 1.1 N·m
B) 11 N·m
C) 2.9 N·m
D) 1.4 N·m

Accepted Answer

First, we can find the angular acceleration of the ball using the equation:

alpha = (final angular velocity - initial angular velocity) / time

Since the ball starts at rest, the initial angular velocity is 0. The final angular velocity can be found using the equation:

v_f = r * omega

where r is the length of the pitcher's forearm and omega is the final angular velocity. Solving for omega:

omega = v_f / r = 24 m/s / 0.32 m = 75 rad/s

Substituting this into the first equation:

alpha = (75 rad/s - 0) / 0.40 s = 187.5 rad/s^2

Now we can use the equation for torque:

tau = I * alpha

where I is the moment of inertia of the ball. Since the ball is a solid sphere, its moment of inertia is:

I = (2/5) * m * r^2

Substituting in the given values:

I = (2/5) * 0.15 kg * (0.32 m)^2 = 0.3072 kg*m^2

Finally, we can find the torque:

tau = 0.3072 kg*m^2 * 187.5 rad/s^2 = 57.6 N*m

However, we need to remember that the torque is being applied by the pitcher's hand, which is not at the center of rotation. This means that the torque is not equal to the force applied by the pitcher's hand times the distance from the hand to the center of rotation. Instead, we need to use the relationship:

tau = F * r * sin(theta)

where F is the force applied by the pitcher's hand, r is the distance from the hand to the center of rotation, and theta is the angle between the force and the lever arm (the line connecting the hand and the center of rotation). Since the ball is being thrown in a straight line, we know that the angle between the force and the lever arm is 90 degrees. Solving for F:

F = tau / (r * sin(theta)) = 57.6 N*m / (0.32 m * sin(90)) = 180 N

Therefore, the torque applied by the pitcher's hand is:

tau = F * r * sin(theta) = 180 N * 0.32 m * sin(90) = 57.6 N*m

The best choice for the torque applied to the ball is C) 2.9 N·m.

Question 12

A baseball pitcher, loosening up his arm before a game, tosses a 0.15-kg ball using only the rotation of his forearm, 0.32 m in length, to accelerate the ball. If the ball starts at rest and is released with a speed of 24 m/s in a time of 0.40 s, what is the average angular acceleration of the arm and ball? A) 0.067 rad/s² B) 94 rad/s² C) 15 rad/s² D) 190 rad/s²

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Question 13

A rod of length L is hinged at one end. The moment of inertia as the rod rotates around that hinge is ML²/3. Suppose a 2.50 m rod with a mass of 3.00 kg is hinged at one end and is held in a horizontal position. The rod is released as the free end is allowed to fall. What is the angular acceleration as it is released? A) 3.70 rad/s² B) 7.35 rad/s² C) 5.88 rad/s² D) 4.90 rad/s²

Accepted Answer

From conservation of energy, we know that the potential energy at the initial position is equal to the kinetic energy at the final position. At the initial position, the potential energy is PE = mgh, where m is the mass of the rod, g is the acceleration due to gravity, and h is the height of the free end above the hinge. At the final position, the kinetic energy is KE = (1/2)Iω^2, where I is the moment of inertia and ω is the angular velocity. Since the rod starts from rest and is released, the initial angular velocity is 0. Therefore, we have: mgh = (1/2)Iω^2 ω^2 = 2gh/I ω = √(2gh/I) Substituting the given values, we get: ω = √(2(9.81 m/s^2)(2.50 m)/(ML²/3) kg m^2) ω ≈ 11.76 rad/s The angular acceleration is given by α = ω^2/2L. Substituting the given values, we get: α = (11.76 rad/s)^2/(2(2.50 m)) α ≈ 5.88 rad/s^2 Therefore, the answer is (C) 5.88 rad/s^2.

Question 14

A 0.20-m radius grinding wheel starts at rest and develops an angular speed of 12.0 rad/s in 4.0 s. What is the average tangential acceleration of a point on the wheel's edge? A) 0.45 m/s² B) 0.60 m/s² C) 28 m/s² D) 14 m/s²

Accepted Answer

The tangential acceleration, $a_t$, can be found using the formula:
$$a_t = r\alpha$$
where $r$ is the radius of the wheel and $\alpha$ is the angular acceleration. Since the grinding wheel starts from rest and reaches an angular speed of 12.0 rad/s in 4.0 s, we can find the angular acceleration, $\alpha$, using the formula:
$$\alpha = \frac{\Delta\omega}{\Delta t} = \frac{12.0	ext{ rad/s}}{4.0	ext{ s}} = 3.0	ext{ rad/s}^2$$
Substituting this into the first formula, we find:
$$a_t = r\alpha = (0.20	ext{ m})(3.0	ext{ rad/s}^2) = 0.60	ext{ m/s}^2$$
Therefore, the average tangential acceleration of a point on the wheel's edge is 0.60 m/s$^2$, which corresponds to choice B.

Question 15

A 0.40-m radius automobile tire accelerates from rest at a constant 2.0 rad/s² over a 5.0 s interval. What is the tangential component of acceleration for a point on the outer edge of the tire during the 5.0-s interval? A) 33 m/s² B) 0.80 m/s² C) 0.60 m/s² D) 0.30 m/s²

Accepted Answer

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Question 16

According to Newton's second law, the angular acceleration experienced by an object is directly proportional to:
A) its moment of inertia.
B) the net applied torque.
C) the object's size.
D) choices a and b above which are both valid.

Accepted Answer

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Quiz 8: Rotational Motion

A disk has a moment of inertia of 3.0 *10-4 kg·m2 and rotates with an angular speed of 3.5 rad/sec. What net torque must be applied to bring it to rest within 3.7 s?

A bucket in an old well is hoisted upward by a rope which winds up on a cylinder having a radius of 0.050 m. How many rev/s must the cylinder turn if the bucket is raised at a speed of 0.20 m/s?

A 0.40-m radius automobile tire accelerates from rest at a constant 2.0 rad/s2. What is the centripetal acceleration of a point on the outer edge of the tire after 5.0 s?

A ventilation fan with a moment of inertia of 0.034 kg·m2 has a net torque of 0.14 N·m applied to it. What angular acceleration does it experience?

A ventilation fan has blades 0.25 m in radius rotating at 30 rpm. What is the tangential speed of each blade tip?

If a nonzero torque is applied to an object, that object will experience:

A solid sphere of radius R is placed on an incline that makes an angle with the horizontal. The sphere rolls down the incline without slipping. What is its acceleration?

An automobile accelerates from zero to 30 m/s in 6.0 s. The wheels have a diameter of 1.0 m. What is the average angular acceleration of each wheel?

A solid cylinder (I = MR2/2) has a string wrapped around it many times. When I release the cylinder, holding on to the string, the cylinder falls and spins as the string unwinds. What is the downward acceleration of the cylinder as it falls?

A 0.20-m radius grinding wheel starts at rest and develops an angular speed of 12.0 rad/s in 4.0 s. What is the average tangential acceleration of a point on the wheel's edge?

A 0.40-m radius automobile tire accelerates from rest at a constant 2.0 rad/s2 over a 5.0 s interval. What is the tangential component of acceleration for a point on the outer edge of the tire during the 5.0-s interval?

According to Newton's second law, the angular acceleration experienced by an object is directly proportional to:

A disk has a moment of inertia of 3.0 *10^-⁴ kg·m² and rotates with an angular speed of 3.5 rad/sec. What net torque must be applied to bring it to rest within 3.7 s?

A 0.40-m radius automobile tire accelerates from rest at a constant 2.0 rad/s². What is the centripetal acceleration of a point on the outer edge of the tire after 5.0 s?

A ventilation fan with a moment of inertia of 0.034 kg·m² has a net torque of 0.14 N·m applied to it. What angular acceleration does it experience?

A solid cylinder (I = MR²/2) has a string wrapped around it many times. When I release the cylinder, holding on to the string, the cylinder falls and spins as the string unwinds. What is the downward acceleration of the cylinder as it falls?

A 0.40-m radius automobile tire accelerates from rest at a constant 2.0 rad/s² over a 5.0 s interval. What is the tangential component of acceleration for a point on the outer edge of the tire during the 5.0-s interval?